eventually approach a closed orbit. We'll discuss this important theorem in Sec-tion 7.3.But that part of our story comes later. First we must become better acquainted

with fixed points.

6.3 Fixed Points and LinearizationIn this section we extend the linearization technique developed earlier for one-dimensional systems (Section 2.4). The hope is that we can approximate the phaseportrait near a fixed point by that of a corresponding linear system.

Linearized System

Consider the system

i=f(x,y)

.v=g(x,y)

and suppose that (x*, y*) is a fixed point, i.e.,

Let

f(x*, y*) =0,

u=x-x*,

g(x*,y*) = O.

v=y-y*

denote the components of a small disturbance from the fixed point. To see whetherthe disturbance grows or decays, we need to deri ve differential equations for u andv. Let's do the u-equation first:

u=x

= f(x *+u, Y *+v)

(since x * is a constant)

(by substitution)

= f(x*, y*) + u af + v af +O(u", v", uv) (Taylor series expansion)ax iJy

af af 1 1= u-+ v- +O(u-, v- ,uv) (since f(x*,y*) = 0).ax iJyTo simplify the notation, we have written allax and all iJy, but please remember thatthese partial derivatives are to be evaluated at thejixed point (x*,y*); thus they arenumbers, not functions. Also, the shorthand notation O(u", v", uv) denotes quadraticterms in II and v. Since u and v small, these quadratic terms are extremely small.Similarly we find

. ag ag 1 1V =11- +v- +O(u-, v- ,uv).at iJy

1SO PHASE PLANE

Hence the disturbance (u, v) evolves according to

( 'J (Jiu a;v =

The matrix

t](:) + quadratic terms.

Ji)a;Jga; Ix*,y*)

(1)

is called the Jacobian matrix at the fixed point (x*, y*). It is the multivariable ana-log of the derivative f'(x*) seen in Section 2.4.Now since the quadratic terms in (1) are tiny, it's tempting to neglect them alto-

gether. Ifwe do that, we obtain the linearized system

(2)

whose dynamics can be analyzed by the methods of Section 5.2.

The Effect of Small Nonlinear Terms

Is it really safe to neglect the quadratic terms in (I)? In other words, does thelinearized system give a qualitatively correct picture of the phase portrait near(x*, y*)? The answer is yes, as long as the fixed point for the linearized systemis not one of the borderline cases discussed in Section 5.2. In other words, ifthe linearized system predicts a saddle, node, or a spiral, then the fixed pointreally is a saddle, node, or spiral for the original nonlinear system. See An-dronov et al. (1973) for a proof of this result, and Example 6.3.1 for a concreteillustration.The borderline cases (centers, degenerate nodes, stars, or non-isolated fixed

points) are much more delicate. They can be altered by small nonlinear terms, aswe'll see in Example 6.3.2 and in Exercise 6.3.11.

EXAMPLE 6.3.1:Find all the fixed points of the system x = -x + x 3 , y = -2y, and use lineariza-

tion to classify them. Then check your conclusions by deriving the phase portraitfor the full nonlinear system.

Solution: Fixed points occur where x= 0 and .v = 0 simultaneously. Hence weneed x = 0 or x = ±I, and y = O. Thus, there are three fixed points: (0,0), (1,0),and (-1,0). The Jacobian matrix at a general point (x,y) is

6.3 FIXED POINTS AND LINEARIZATION 151

J (-1 +3x2A = i), aj =ax ay 0

Next we evaluate A at the fixed points. At (0,0) , we find A =(-1 0 ), so. 0 -2

(0,0) is a stable node. At (±1,0), A = (2 0), so both (1,0) and (-1,0) are sad-o -2dIe points.Now because stable nodes and saddle points are not borderline cases, we can be

certain that the fixed points for the full nonlinear system have been predicted cor-rectly.This conclusion can be checked explicitly for the nonlinear system, since the

x and y equations are uncoupled; the system is essentially two independentfirst-order systems at right angles to each other. In the y-direction, all trajecto-ries decay exponentially to y = O. In the x-direction, the trajectories are attractedto x = 0 and repelled from x = ±l. The vertical lines x = 0 and x = ±l are in-variant, because x= 0 on them; hence any trajectory that starts on these linesstays on them forever. Similarly, y = 0 is an invariant horizontal line. As a finalobservation, we note that the phase portrait must be symmetric in both the x andy axes, since the equations are invariant under the transformations x -x andy -yo Putting all this information together, we arrive at the phase portraitshown in Figure 6.3.1.

y lx

\ IFigure 6.3.1

This picture confirms that (0,0) is a stable node, and (±1,0) are saddles, as ex-pected from the linearization. _

The next example shows that small nonlinear terms can change a center into a spiral.

152 PHASE PLANE

EXAMPLE 6.3.2:Consider the system

i=-y+ax(x2+/)

y = x + ay(x 2 + /)

where a is a parameter. Show that the linearized system incorrectly predicts thatthe origin is a center for all values of a, whereas in fact the origin is a stable spiralif a < 0 and an unstable spiral if a> O.

Solution: To obtain the linearization about (x*, y*) = (0,0), we can either com-pute the Jacobian matrix directly from the definition, or we can take the followingshortcut. For any system with a fixed point at the origin, x and y represent devia-tions from the fixed point, since u = x - x* = x and v = y - y* = y;. hence we canlinearize by simply omitting nonlinear terms in x and y . Thus the linearized sys-tem is i =-y, y=x. The Jacobian is

which has r =0 , ,1 =1> 0, so the origin is always a center, according to the lin-earization.To analyze the nonlinear system, we change variables to polar coordinates. Let

x = r cos 0, y = r sin G. To derive a differential equation for r, we note xZ + / = r Z,so xi +yy = rr. Substituting for i and y yields

rr = x(-y + ax(x z + i)) + y(x + ay(x z + i))= a(xZ + /)z=ar4 •

Hence r= ar3• In Exercise 6.3.12, you are asked to derive the following differen-tial equation for 0 :

. xy- yxo=--z-·

r

After substituting for i and y we find iJ =1. Thus in polar coordinates the originalsystem becomes

. 3r=ar

The system is easy to analyze in this form, because the radial and angular mo-

6.3 FIXED POINTS AND LINEARIZATION 153

tions are independent. All trajectories rotate about the origin with constant angularvelocity f) =1.The radial motion depends on a, as shown in Figure 6.3.2.

, ,,,,,,,

a<O a=O

,,,, ,,,,

a>O

Figure 6.3.2

If a < 0 , then ret) 0 monotonically as t 00 • In this case, the origin is a sta-ble spiral. (However, note that the decay is extremely slow, as suggested by thecomputer-generated trajectories shown in Figure 6.3.2.) If a = 0, then ret) = rofor all t and the origin is a center. Finally, if a> 0, then ret) 00 monotonicallyand the origin is an unstable spiral.We can see now why centers are so delicate: all trajectories are required to close

perfectly after one cycle. The slightest miss converts the center into a spiral. _

Similarly, stars and degenerate nodes can be altered by small nonlinearities, butunlike centers, their stability doesn't change. For example, a stable star may bechanged into a stable spiral (Exercise 6.3.11) but not into an unstable spiral. This isplausible, given the classification of linear systems in Figure 5.2.8: stars and de-generate nodes live squarely in the stable or unstable region, whereas centers liveon the razor's edge between stability and instability.If we're only interested in stability, and not in the detailed geometry of the tra-

jectories, then we can classify fixed points more coarsely as follows:

Robust cases:Repellers (also called sources): both eigenvalues have positive realpart.

Attractors (also called sinks): both eigenvalues have negative real part.Saddles: one eigenvalue is positive and one is negative.

Marginal cases:Centers: both eigenvalues are pure imaginary.Higher-order and non-isolated fixed points: at least one eigenvalue iszero.

Thus, from the point of view of stability, the marginal cases are those where atleast one eigenvalue satisfies Re(A.-) =o.

154 PHASE PLANE

Hyperbolic Fixed Points, Topological Equivalence, andStructural Stability

If Re(A)"* 0 for both eigenvalues, the fixed point is often called hyperbolic.(This is an unfortunate name-it sounds like it should mean "saddle point"-but ithas become standard.) Hyperbolic fixed points are sturdy; their stability type is un-affected by small nonlinear terms. Nonhyperbolic fixed points are the fragile ones.We've already seen a simple instance of hyperbolicity in the context of vector

fields on the line. In Section 2.4 we saw that the stability of a fixed point was accu-rately predicted by the linearization, as long as f'(x*)"* O. This condition is theexact analog of Re(A)"* O.These ideas also generalize neatly to higher-order systems. A fixed point of an

nth-order system is hyperbolic if all the eigenvalues of the linearization lie off theimaginary axis, i.e., Re(A,)"* 0 for i =I, ... , n. The important Hartman-Grobman theorem states that the local phase portrait near a hyperbolic fixed pointis "topologically equivalent" to the phase portrait of the linearization; in particular,the stability type of the fixed point is faithfully captured by the linearization. Heretopologically equivalent means that there is a homeomorphism (a continuous de-formation with a continuous inverse) that maps one local phase portrait onto theother, such that trajectories map onto trajectories and the sense of time (the direc-tion of the arrows) is preserved.Intuitively, two phase portraits are topologically equivalent if one is a distorted

version of the other. Bending and warping are allowed, but not ripping, so closed or-bits must remain closed, trajectories connecting saddle points must not be broken, etc.Hyperbolic fixed points also illustrate the important general notion of structural

stability. A phase portrait is structurally stable if its topology cannot be changedby an arbitrarily small perturbation to the vector field. For instance, the phase por-trait of a saddle point is structurally stable, but that of a center is not: an arbitrarilysmall amount of damping converts the center to a spiral.

6.4 Rabbits versus SheepIn the next few sections we'll consider some simple examples of phase planeanalysis. We begin with the classic Lotka-Volterra model ofcompetition betweentwo species, here imagined to be rabbits and sheep. Suppose that both species arecompeting for the same food supply (grass) and the amount available is limited.Furthermore, ignore all other complications, like predators, seasonal effects, andother sources of food. Then there are two main effects we should consider:

1. Each species would grow to its carrying capacity in the absence of theother. This can be modeled by assuming logistic growth for eachspecies (recall Section 2.3). Rabbits have a legendary ability to repro-duce, so perhaps we should assign them a higher intrinsic growth rate.

6.4 RABBITS VERSUS SHEEP 155

2. When rabbits and sheep encounter each other, trouble starts. Some-times the rabbit gets to eat, but more usually the sheep nudges therabbit aside and starts nibbling (on the grass, that is). We'll assumethat these conflicts occur at a rate proportional to the size of eachpopulation. (If there were twice as many sheep, the odds of a rabbitencountering a sheep would be twice as great.) Furthermore, we as-sume that the conflicts reduce the growth rate for each species, butthe effect is more severe for the rabbits.

A specific model that incorporates these assumptions is

x= x(3 - x - 2y)Y= y(2-x- y)

where

x(t) = population of rabbits,yet) = population of sheep

and x, y O. The coefficients have been chosen to reflect this scenario, but are oth-.erwise arbitrary. In the exercises, you'll be asked to study what happens if the co-efficients are changed.To find the fixed points for the system, we solve x= 0 and y= 0 simultane-

ously. Four fixed points are obtained: (0,0), (0,2), (3,0), and (1,1). To classifythem, we compute the Jacobian:

[il" il") (....::.....::. 3-2x-2yA_ax ay_

- - -y-2x )

2-X'-2y .

Now consider the four fixed points in tum:

(0,0): Then A =

The eigenvalues are A= 3, 2 so (0,0) is an unstable node. Trajectories leavethe origin parallel to the eigenvector for A= 2 , i.e. tangential toYL v = (0,1), which spans the y-axis. (Recall the general rule: at anode, trajectories are tangential to the slow eigendirection,which is the eigendirection with the smallest IAI.) Thus, thephase portrait near (0,0) looks like Figure 6.4.1.

Figure 6.4.1

x

(-1 0)(0,2): Then A = .-2 -2

This matrix has eigenvalues A = -1, -2 , as can be seen from inspection, since

156 PHASE PLANE

the matrix is triangular. Hence the fixed point is a stable node. Trajectories ap-proach along the eigendirection associated with A. = -1 ; you can check that this di-rection is spanned by v = (1, -2). Figure 6.4.2 shows the phase portrait near thefixed point (0,2).

x

Figure 6.4.2

(3,0): Then A = and A. = -3,-1.

This is also a stable node. The trajectories approach along the slow eigendirec-tion spanned by v = (3, -1), as shown in Figure 6.4.3.

Figure 6.4.3

(1,1): Then A=(-l -2), which has r=-2, L1=-1, and A.=-l±-.!2.-1 -1

Hence this is a saddle point. As you can check, the phase portrait near (1,1) is asshown in Figure 6.4.4.

x

Figure 6.4.4

Combining Figures 6.4.1-6.4.4, we get Figure 6.4.5, which already conveys agood sense of the entire phase portrait. Furthermore, notice that the x and y axescontain straight-line trajectories, since x= 0 when x = 0, and y= 0 when y= o.

6.4 RABBITS VERSUS SHEEP 157

x

Figure 6.4.5

Now we use common sense to fill in the rest of the phase portrait (Figure 6.4.6).For example, some of the trajectories starting near the origin must go to the stablenode on the x-axis, while others must go to the stable node on the y-axis. In be-tween, there must be a special trajectory that can't decide which way to turn, andso it dives into the saddle point. This trajectory is part of the stable manifold of thesaddle, drawn with a heavy line in Figure 6.4.6.

y

x

Figure 6.4.6

32

2

sheep

Figure 6.4.7

The other branch of the stable manifuld consists of a trajectory coming in "from in-finity." A computer-generated phase portrait (Figure 6.4.7) confirms our sketch.

The phase portrait has an inter-esting biological interpretation. Itshows that one species generallydrives the other to extinction. Tra-jectories starting below the stablemanifold lead to eventual extinc-tion of the sheep, while those start-ing above lead to eventualextinction of the rabbits. This di-

rabbits chotomy occurs in other models ofcompetition and has led biologiststo formulate the principle of com-

petitive exclusion, which states that two species competing for the same limited re-source typically cannot coexist. See Pianka (1981) for a biological discussion, and

158 PHASE PLANE

Pielou (1969), Edelstein-Keshet (1988), or Murray (1989) for additional refer-ences and analysis.Our example also illustrates some general mathematical concepts. Given an at-

tracting fixed point x *,we define its basin ofattraction to be the set of initial con-ditions X o such that xU) -7 x * as t -7 00. For instance, the basin of attraction forthe node at (3,0) consists of all the points lying below the stable manifold of thesaddle. This basin is shown as the shaded region in Figure 6.4.8.

basin boundary =stable manifold of saddle

sheep

2

2 3

Figure 6.4.8

Because the stable manifold separates the basins for the two nodes, it is called thebasin boundary. For the same reason, the two trajectories that comprise the stablemanifold are traditionally called separatrices. Basins and their boundaries are im-portant because they partition the phase space into regions of different long-termbehavior.

6.5 Conservative SystemsNewton's law F = rna is the source of many important second-order systems. Forexample, consider a particle of mass rn moving along the x-axis, subject to a non-linear force F(x). Then the equation of motion is

mi =F(x).Notice that we are assuming that F is independent of both x and t ; hence there isno damping or friction of any kind, and there is no time-dependent driving force.Under these assumptions, we can show that energy is conserved, as follows. Let

Vex) denote the potential energy, defined by F(x) = -dV/dx. Then

.. dV 0rnx+-= . (1)dx

6.5 CONSERVATIVE SYSTEMS 159